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# Constant of Integration

Top
 Sub Topics Since the derivative of a constant is zero, any constant may be added to an indefinite integral (i.e., antiderivative) and will still correspond to the same integral. Another way of stating this is that the antiderivative is a nonunique inverse of the derivative. For this reason, indefinite integrals are often written in the form:$\int$ f(x) dx = F(x) + Cwhere C is an arbitrary constant known as the constant of integration.

## constant of integration definition

In calculus, the indefinite integral of a given function is always written with a constant, the constant of integration. This constant expresses an ambiguity inherent in the construction of antiderivatives. If a function f(x) is defined on an interval and F(x) is an antiderivative of f(x), then the set of all antiderivatives of f(x) is given by the functions F(x) + C, where C is an arbitrary constant.

## constant of integration examples

Given below are some examples.

Example 1: $\int$ $\frac{x^{2}+ 1}{(1 + x)^{2}}$ $e^{x}$ $dx$

Solution: $\int$ $e^{x}$ $[ \frac{(x + 1)^{2} - 2x}{(x + 1)^{2}}]$ + $dx$

$\int$ $e^{x}$ [ 1 - $\frac{2x }{(x + 1)^{2}}$] $dx$

$\int$ $e^{x}dx$ - 2 $\int e^{x}$ $\frac{x}{(x + 1)^{2}}$ $dx$

$e^{x}$ - 2$\int e^{x}$ $[\frac{x +1}{(x + 1)^{2}} - \frac{1}{(x + 1)^{2}}]$ $dx$

e$^{x}$-2$\int e^{x}$ $[\frac{1}{x + 1} + \frac{-1}{(x + 1)^{2}}]$

$e^{x}$ - 2 $e^{x}$($\frac{1}{x + 1}$) + $c$

$e^{x}$(1 - $\frac{2}{x + 1}$) + $c$

= $e^{x}$($\frac{x - 1}{x + 1}$) + $c$

Example 2: Integrate $\frac{(log x)^{n}}{x}$  with respect to x

Solution : Let I = $\int$ $\frac{(log x)^{n}}{x}$ $dx$

Put $log x$ = $t$ then, $\frac{1}{x}$ $dx = dt$

Therefore I = $\int t^{n}$ $dt$

= $\frac{t^{n + 1}}{n + 1}$ + $c$

= $\frac{( log x)^{n + 1}}{n + 1}$ + c

Example 3: Integrate $\frac{1}{a^{2} - x^{2}}$ w.r.t x

Solution: Take I = $\int$ $\frac{1}{a^{2}- x^{2}}$ $dx$

$\frac{1}{a^{2}- x^{2}}$ = $\frac{A}{a - x}$ + $\frac{B}{a + x}$

I = A(a + x) + B(a - x)

Put x = a, we get I= 2aA

A = $\frac{1}{2a}$

Put x = - a, we get I = 2Ba

B = $\frac{1}{2a}$

I = $\int$ $\frac{1}{2a}\frac{dx}{(a- x)}$ + $\int$ $\frac{1}{2a}$$\frac{dx}{a+x}$

= - $\frac{1}{2a}$ $log (a - x)$+ $\frac{1}{2a}$ $log(a+ x) + c$

= $\frac{1}{2a}$ $log$($\frac{a+x}{a-x}$) + C

Example 4: $\int_{0}^{1}xe^{x^{2}}$dx

Solution: Put x$^{2}$ = t

when x = 0, 2x $dx = dt$

t = 0

xdx = $\frac{dt}{2}$

When x  = 1, $\int_{0}^{1}e^{t}$ $\frac{dt}{2}$

= $\frac{1}{2}$ $\int_{0}^{1}e^{t}dt$

t = 1 = $\frac{1}{2}$ $e^{t}]_{0}^{1}$

= $\frac{1}{2}$ $[e^{t} - e^{0}]$

= $\frac{1}{2}$ $[e - 1]$

Example  5: If the marginal revenue is 3$x^{2}$ - 4$x$ + 3, find the total revenue and average revenue where x is the output. Write the demand function.

Solution: Marginal revenue 3$x^{2}$ - 4$x$ + 3

Total revenue [ 3 $\int x^{2}$ - 4$\int$ x + 3]dx

Average Revenue = $\frac{Total\ \ Revenue}{x}$

$\frac{3x^{3}}{3}$  - $\frac{4x^{2}}{2}$ + 3x

= $x^{3}$ - 2$x^{2}$ + 3x

Average Revenue = Demand function = $\frac{x^{3}-2x^{2}+3x}{x}$

= x$^{2}$ - 2x + 3

Example 6: The marginal cost function f'(C) = 1 - 2x + 24x$^{3}$ where x is the output. Find the total cost, average cost, variable cost and average variable cost, given that fixed cost is 70.

Solution: Fixed cost = 70

Total cost = $\int$ Marginal cost dx

= $\int$ 1 - 2x + 24x$^{3}$ $dx$

= x - $\frac{2x^{2}}{2}$ + $\frac{24x^{3}}{3}$ + C

Total cost = $x$ - $x^{2}$ + 8$x^{3}$ + 70

Total variable cost = $x$ - $x^{2}$ + 8$x^{3}$

Average cost = $\frac{Total\ \ cost}{x}$

= 1 - x + 8$x^{2}$ + $\frac{70}{x}$

Average variable cost = 1 - x + 8$x^{2}$